What is the tangent of y=8+4x^2-2x^3,(2,8) ?

The tangent of y=8+4x^2-2x^3,(2,8) is y=-8x+24

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tangent of y=8+4x^2-2x^3,(2,8)

tangent of

y = 8 + 4 x^{2} - 2 x^{3}, (2, 8)

y = - 8 x + 24

Solution steps

Find the slope of

y = 8 + 4 x^{2} - 2 x^{3} : \frac{d y}{d x} = - 6 x^{2} + 8 x

Compute the slope of

y = 8 + 4 x^{2} - 2 x^{3}

(2, 8) : m = - 8

Find the line with slope m=

- 8

and passing through

(2, 8) : y = - 8 x + 24

y = - 8 x + 24

x \to 2 lim (x^{2} + 3 x - 4)

\int \frac{x}{x ^{4} + 7} d x

\int (x^{2} + y^{2} + z^{2}) d x

\int \frac{3 + 6 x}{1 + x ^{2}} d x

\frac{d}{d x} (x \cdot ln (x + 1))

What is the tangent of y=8+4x^2-2x^3,(2,8) ?
The tangent of y=8+4x^2-2x^3,(2,8) is y=-8x+24